Count of subsets having sum of min and max element less than K

Given an integer array arr[] and an integer K, the task is to find the number of non-empty subsets S such that min(S) + max(S) < K.
Examples: 
 

Input: arr[] = {2, 4, 5, 7} K = 8 
Output: 4 
Explanation: 
The possible subsets are {2}, {2, 4}, {2, 4, 5} and {2, 5}
Input:: arr[] = {2, 4, 2, 5, 7} K = 10 
Output: 26 
 

Approach 

• Sort the input array first.
• Now use Two Pointer Technique to count the number of subsets.
• Let take two pointers left and right and set left = 0 and right = N-1.

if (arr[left] + arr[right] < K ) 
Increment the left pointer by 1 and add 2 ^{j - i} into answer, because the left and right values make up a potential end values of a subset. All the values from [i, j - 1] also make up end of subsets which will have the sum < K. So, we need to calculate all the possible subsets for left = i and right ? [i, j]. So, after summing up values 2 ^{j - i + 1} + 2 ^{j - i - 2} + ... + 2 ⁰ of the GP, we get 2 ^{j - i} . 
if( arr[left] + arr[right] >= K ) 
Decrement the right pointer by 1. 
 

• Repeat the below process until left <= right.

Below is the implementation of the above approach:

// C++ program to print count
// of subsets S such that
// min(S) + max(S) < K

#include <bits/stdc++.h>
using namespace std;

// Function that return the
// count of subset such that
// min(S) + max(S) < K
int get_subset_count(int arr[], int K,
                     int N)
{
    // Sorting the array
    sort(arr, arr + N);

    int left, right;
    left = 0;
    right = N - 1;

    // ans stores total number of subsets
    int ans = 0;

    while (left <= right) {
        if (arr[left] + arr[right] < K) {

            // add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else {
            // Decrease the sum
            right--;
        }
    }
    return ans;
}

// Driver code
int main()
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << get_subset_count(arr, K, N);
    return 0;
}

// Java program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K
import java.util.*;

class GFG{

// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
static int get_subset_count(int arr[], int K,
                                       int N)
{
    
    // Sorting the array
    Arrays.sort(arr);

    int left, right;
    left = 0;
    right = N - 1;

    // ans stores total number
    // of subsets
    int ans = 0;

    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {

            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else 
        {
            
            // Decrease the sum
            right--;
        }
    }
    return ans;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.length;
    
    System.out.print(get_subset_count(arr, K, N));
}
}

// This code is contributed by Rajput-Ji

# Python3 program to print 
# count of subsets S such 
# that min(S) + max(S) < K 

# Function that return the
# count of subset such that
# min(S) + max(S) < K 
def get_subset_count(arr, K, N):

    # Sorting the array 
    arr.sort() 

    left = 0; 
    right = N - 1; 

    # ans stores total number of subsets 
    ans = 0; 

    while (left <= right):
        if (arr[left] + arr[right] < K):
            
            # Add all possible subsets 
            # between i and j 
            ans += 1 << (right - left); 
            left += 1; 
        else:
            
            # Decrease the sum 
            right -= 1; 
    
    return ans; 

# Driver code 
arr = [ 2, 4, 5, 7 ]; 
K = 8; 

print(get_subset_count(arr, K, 4))

# This code is contributed by grand_master

// C# program to print count
// of subsets S such that
// Math.Min(S) + Math.Max(S) < K
using System;

class GFG{

// Function that return the
// count of subset such that
// Math.Min(S) + Math.Max(S) < K
static int get_subset_count(int []arr, int K,
                                       int N)
{
    
    // Sorting the array
    Array.Sort(arr);

    int left, right;
    left = 0;
    right = N - 1;

    // ans stores total number
    // of subsets
    int ans = 0;

    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
            
            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
            
            // Decrease the sum
            right--;
        }
    }
    return ans;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.Length;
    
    Console.Write(get_subset_count(arr, K, N));
}
}

// This code is contributed by gauravrajput1

<script>

// JavaScript program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K

// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
function get_subset_count(arr,K,N)
{
    // Sorting the array
    (arr).sort(function(a,b){return a-b;});
 
    let left, right;
    left = 0;
    right = N - 1;
 
    // ans stores total number
    // of subsets
    let ans = 0;
 
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
 
            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
             
            // Decrease the sum
            right--;
        }
    }
    return ans;
}

// Driver code
let arr=[ 2, 4, 5, 7];
let K = 8;
let N = arr.length;
document.write(get_subset_count(arr, K, N));


// This code is contributed by patel2127

</script>

4

Time Complexity: O(N* log N) 
Auxiliary Space: O(1)